Mathematics
redraven
37

What is the 4-digit number in which the first digit is one-fifth of the last, and the second and third digits are the last digit multiplied by three?

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(1) Answers
Mando

We'll call this number [latex]A B C D[/latex] We have [latex]A=\frac{1}{5}D[/latex]⇒[latex]D=5A[/latex] Since [latex]D[/latex] is divisible by 5, it is either 0 or 5 But it can't be 0, because then [latex]A[/latex] would be 0 So [latex]D=5[/latex] and [latex]A=1[/latex] Now I'm assuming that the second and third digits are adding to three times the last digit. So [latex]B+C=3*5[/latex]⇒[latex]B+C=15[/latex] We have a number of options: [latex](B, C)=(6, 9)[/latex] [latex](B, C)=(7, 8)[/latex] [latex](B, C)=(8, 7)[/latex] [latex](B, C)=(9, 6)[/latex] So the possibilities are: 1695 1785 1875 1965 (If there were another detail we could narrow down the choices further, but otherwise we have all these options.)

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